\(\int (d+e x) (c d^2+2 c d e x+c e^2 x^2)^2 \, dx\) [988]

Optimal result

Integrand size = 28, antiderivative size = 17 \[ \int (d+e x) \left (c d^2+2 c d e x+c e^2 x^2\right )^2 \, dx=\frac {c^2 (d+e x)^6}{6 e} \]

[Out]

1/6*c^2*(e*x+d)^6/e

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.107, Rules used = {27, 12, 32} \[ \int (d+e x) \left (c d^2+2 c d e x+c e^2 x^2\right )^2 \, dx=\frac {c^2 (d+e x)^6}{6 e} \]

[In]

Int[(d + e*x)*(c*d^2 + 2*c*d*e*x + c*e^2*x^2)^2,x]

[Out]

(c^2*(d + e*x)^6)/(6*e)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rubi steps \begin{align*} \text {integral}& = \int c^2 (d+e x)^5 \, dx \\ & = c^2 \int (d+e x)^5 \, dx \\ & = \frac {c^2 (d+e x)^6}{6 e} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00 \[ \int (d+e x) \left (c d^2+2 c d e x+c e^2 x^2\right )^2 \, dx=\frac {c^2 (d+e x)^6}{6 e} \]

[In]

Integrate[(d + e*x)*(c*d^2 + 2*c*d*e*x + c*e^2*x^2)^2,x]

[Out]

(c^2*(d + e*x)^6)/(6*e)

Maple [A] (verified)

Time = 2.37 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.82

method	result	size
default	\(\frac {\left (c \,x^{2} e^{2}+2 x c d e +c \,d^{2}\right )^{3}}{6 c e}\)	\(31\)
gosper	\(\frac {x \left (e^{5} x^{5}+6 x^{4} d \,e^{4}+15 d^{2} e^{3} x^{3}+20 d^{3} e^{2} x^{2}+15 d^{4} e x +6 d^{5}\right ) c^{2}}{6}\)	\(58\)
norman	\(d^{5} c^{2} x +c^{2} d \,x^{5} e^{4}+\frac {1}{6} c^{2} x^{6} e^{5}+\frac {5}{2} c^{2} d^{2} e^{3} x^{4}+\frac {10}{3} c^{2} d^{3} e^{2} x^{3}+\frac {5}{2} d^{4} e \,c^{2} x^{2}\)	\(72\)
parallelrisch	\(d^{5} c^{2} x +c^{2} d \,x^{5} e^{4}+\frac {1}{6} c^{2} x^{6} e^{5}+\frac {5}{2} c^{2} d^{2} e^{3} x^{4}+\frac {10}{3} c^{2} d^{3} e^{2} x^{3}+\frac {5}{2} d^{4} e \,c^{2} x^{2}\)	\(72\)
risch	\(\frac {c^{2} x^{6} e^{5}}{6}+c^{2} d \,x^{5} e^{4}+\frac {5 c^{2} d^{2} e^{3} x^{4}}{2}+\frac {10 c^{2} d^{3} e^{2} x^{3}}{3}+\frac {5 d^{4} e \,c^{2} x^{2}}{2}+d^{5} c^{2} x +\frac {c^{2} d^{6}}{6 e}\)	\(83\)

[In]

int((e*x+d)*(c*e^2*x^2+2*c*d*e*x+c*d^2)^2,x,method=_RETURNVERBOSE)

[Out]

1/6*(c*e^2*x^2+2*c*d*e*x+c*d^2)^3/c/e

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 71 vs. \(2 (15) = 30\).

Time = 0.27 (sec) , antiderivative size = 71, normalized size of antiderivative = 4.18 \[ \int (d+e x) \left (c d^2+2 c d e x+c e^2 x^2\right )^2 \, dx=\frac {1}{6} \, c^{2} e^{5} x^{6} + c^{2} d e^{4} x^{5} + \frac {5}{2} \, c^{2} d^{2} e^{3} x^{4} + \frac {10}{3} \, c^{2} d^{3} e^{2} x^{3} + \frac {5}{2} \, c^{2} d^{4} e x^{2} + c^{2} d^{5} x \]

[In]

integrate((e*x+d)*(c*e^2*x^2+2*c*d*e*x+c*d^2)^2,x, algorithm="fricas")

[Out]

1/6*c^2*e^5*x^6 + c^2*d*e^4*x^5 + 5/2*c^2*d^2*e^3*x^4 + 10/3*c^2*d^3*e^2*x^3 + 5/2*c^2*d^4*e*x^2 + c^2*d^5*x

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 80 vs. \(2 (12) = 24\).

Time = 0.02 (sec) , antiderivative size = 80, normalized size of antiderivative = 4.71 \[ \int (d+e x) \left (c d^2+2 c d e x+c e^2 x^2\right )^2 \, dx=c^{2} d^{5} x + \frac {5 c^{2} d^{4} e x^{2}}{2} + \frac {10 c^{2} d^{3} e^{2} x^{3}}{3} + \frac {5 c^{2} d^{2} e^{3} x^{4}}{2} + c^{2} d e^{4} x^{5} + \frac {c^{2} e^{5} x^{6}}{6} \]

[In]

integrate((e*x+d)*(c*e**2*x**2+2*c*d*e*x+c*d**2)**2,x)

[Out]

c**2*d**5*x + 5*c**2*d**4*e*x**2/2 + 10*c**2*d**3*e**2*x**3/3 + 5*c**2*d**2*e**3*x**4/2 + c**2*d*e**4*x**5 + c
**2*e**5*x**6/6

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.76 \[ \int (d+e x) \left (c d^2+2 c d e x+c e^2 x^2\right )^2 \, dx=\frac {{\left (c e^{2} x^{2} + 2 \, c d e x + c d^{2}\right )}^{3}}{6 \, c e} \]

[In]

integrate((e*x+d)*(c*e^2*x^2+2*c*d*e*x+c*d^2)^2,x, algorithm="maxima")

[Out]

1/6*(c*e^2*x^2 + 2*c*d*e*x + c*d^2)^3/(c*e)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 60 vs. \(2 (15) = 30\).

Time = 0.28 (sec) , antiderivative size = 60, normalized size of antiderivative = 3.53 \[ \int (d+e x) \left (c d^2+2 c d e x+c e^2 x^2\right )^2 \, dx=\frac {1}{2} \, {\left (e x^{2} + 2 \, d x\right )} c^{2} d^{4} + \frac {1}{2} \, {\left (e x^{2} + 2 \, d x\right )}^{2} c^{2} d^{2} e + \frac {1}{6} \, {\left (e x^{2} + 2 \, d x\right )}^{3} c^{2} e^{2} \]

[In]

integrate((e*x+d)*(c*e^2*x^2+2*c*d*e*x+c*d^2)^2,x, algorithm="giac")

[Out]

1/2*(e*x^2 + 2*d*x)*c^2*d^4 + 1/2*(e*x^2 + 2*d*x)^2*c^2*d^2*e + 1/6*(e*x^2 + 2*d*x)^3*c^2*e^2

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 71, normalized size of antiderivative = 4.18 \[ \int (d+e x) \left (c d^2+2 c d e x+c e^2 x^2\right )^2 \, dx=c^2\,d^5\,x+\frac {5\,c^2\,d^4\,e\,x^2}{2}+\frac {10\,c^2\,d^3\,e^2\,x^3}{3}+\frac {5\,c^2\,d^2\,e^3\,x^4}{2}+c^2\,d\,e^4\,x^5+\frac {c^2\,e^5\,x^6}{6} \]

[In]

int((d + e*x)*(c*d^2 + c*e^2*x^2 + 2*c*d*e*x)^2,x)

[Out]

c^2*d^5*x + (c^2*e^5*x^6)/6 + (5*c^2*d^4*e*x^2)/2 + c^2*d*e^4*x^5 + (10*c^2*d^3*e^2*x^3)/3 + (5*c^2*d^2*e^3*x^
4)/2